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Homework Help: Integration by Parts - Choice of variables

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm getting different results when choosing my u & dv for Integration by Parts on the following integral:

    [tex]\int 2x^3 e^x^2 dx [/tex]
    (Note, the exponent on 'e' is x^2)

    This yields the correct solution:
    u = [tex]x^2[/tex]
    dv = [tex]2x e^x^2 dx [/tex]

    du = [tex]2xdx[/tex]
    v = [tex]e^x^2[/tex]

    However, I have tried using this instead (*)

    u = [tex]2x^3[/tex]
    dv = [tex]e^x^2 dx [/tex]

    du = [tex]6x^2 dx[/tex]
    v = [tex](e^x^2) / 2x[/tex]

    and this is yielding the incorrect solution (see 3.)

    2. Relevant equations
    Integration by Parts:
    [tex]\int udv = uv - \int vdu [/tex]


    3. The attempt at a solution

    The correct solution turns out to be
    [tex] x^2 e^x^2 - e^x^2 + C[/tex]

    When I use my other choice of variables (*), I get (using IBP)
    [tex] \int 2x^3 e^x^2 dx = 2x^3 e^x^2 / 2x - \int e^x^2 / 2x * 6x^2 dx [/tex]
    which leads to:
    [tex]x^2 e^x^2 - 3/2 e^x^2 + C [/tex]

    which is different from the other choice of variables.

    I've looked over both choices of variables, and I don't know why the second choice (*) comes up with a different solution.

    Thanks for the help!
     
  2. jcsd
  3. Feb 19, 2010 #2

    vela

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    Your error is that the integral of [tex]e^{x^2}[/tex] isn't [tex]e^{x^2}/(2x)[/tex]. (I'm assuming you meant to have parentheses on the bottom, but perhaps not. Regardless, the answer is incorrect either way.)
     
  4. Feb 19, 2010 #3
    Ah! Silly me, I've been staring at it and completely overlooked that. Thanks for your quick reply, I'll avoid such carelessness in the future :)
     
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